题意:给你n,m分别表示 长度为n的环 和最后走到的位置m 问从0点出发每次都能能往前或者往后走 求最后在m点的概率
思路:我们可以先打表模拟一下 发现好像每个点的概率大概都是1/(n-1)
打表代码:
#include<bits/stdc++.h> #include <random> #include <chrono> #define ll long long #define ull unsigned long long const int inf = 0x3f3f3f3f; const int N = 4e5+7; const ll mod = 998244353; using namespace std; int t,now,n,x; int vis[105],book[105]; int main(){ cin>>n; unsigned seed = chrono::high_resolution_clock::now().time_since_epoch().count() ; mt19937 rand_generator(seed) ; uniform_int_distribution<int> dist(0,1); for(int i=1;i<=100000;i++){ memset(vis,0,sizeof(vis)); vis[0] = 1; now = 0; t = 1; while(t<n){ x=dist(rand_generator)?1:-1; now+=x; now=(now+n)% n; if(!vis[now]){ vis[now]=1; t++; } if(t==n) book[now]++; } } for(int i=0;i<n;i++) cout<<book[i]*1.0/100000<<' '; cout<<endl; }
ac代码:
#include<bits/stdc++.h> #define ll long long int using namespace std; const int inf=0x3f3f3f3f; const ll mod=1e9+7; const int N = 1e5+7; ll q_pow(ll a,ll n){ ll ans=1; ll base=a; while(n){ if(n&1) ans=(ans*base)%mod; base=base*base%mod; n>>=1; } return ans; } ll inv(ll a,ll b){ return q_pow(a,b-2); } int main(){ ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); int t; cin>>t; ll ans=1; while(t--){ ll n,m; cin>>n>>m; if(m==0){ if(n>1) ans=0; } else{ ans=(ans%mod*inv(n-1,mod)%mod)%mod; } cout<<ans<<endl; } return 0; }